 Solution

Both assertions are false. For simple closure, take the inclusionBoth satisfy DAG prime, but the quantifier-free formula $x\ne 0$ has a witness in $N$ and none in $M$. The same example works in the reduced language $\{+,0\}$.

For quantifier elimination, the first-order sentence $\exists x(x\ne 0)$ distinguishes these two models. Every closed group term is zero, so every atomic closed equality is true in both models. Every quantifier-free sentence, being a Boolean combination of such equalities, has the same truth value in both. The distinguishing sentence therefore has no equivalent quantifier-free sentence modulo DAG prime. HenceIncluding the trivial group is exactly what makes the proposed simple closure condition fail.